Question
What volume of 0.1 M NaOH solution is required to neutralise 100 ml of concentrated aqueous sulphuric acid which contains $98 \% \mathrm{~H}_2 \mathrm{SO}_4$ by mass. The density of concentrated sulphuric acid solution is $1.84 \mathrm{g} \mathrm{ml}^{-1} \mathrm{NaOH}$ reacts with $\mathrm{H}_2 \mathrm{SO}_4$ according to the following reaction:
$2 \mathrm{NaOH}+\mathrm{H}_2 \mathrm{SO} 4 \rightarrow \mathrm{Na}_2 \mathrm{SO}_4+2 \mathrm{H}_2 \mathrm{O}$
(Atomic mass $/ \mathrm{g} \mathrm{mol}^{-1} \mathrm{H}=1, \mathrm{~S}=32, \mathrm{O}=16$ ).

Answer

Molarity $=\frac{\text{Number of moles of solute}}{\text{Litre of solution}}$

$=\frac{\text{Mass of solute}\big(\text{W}_{\text{B}}\big)}{\text{Molar mass of solute}\big(\text{M}_{\text{B}}\big)}\times\frac{1000}{\text{Volume of solution in mL}}$

$=\frac{\text{Mass of solute}\big(\text{W}_{\text{B}}\big)}{\text{Molar mass of solute}\big(\text{M}_{\text{B}}\big)}\times\frac{1000}{\frac{\text{Volume of solution}}{\text{Density of solution}}}$

$=\frac{98}{98}\times\frac{1000}{\frac{\text{Volume of solution}}{\text{Density of solution}}}$

$=\frac{98}{98}\times\frac{1000}{\frac{100}{1.84}}=18.4\text{M}$

$=\frac{98}{98}\times\frac{1000}{100}\times1.84$

$2(\text{M}_1\text{V}_1)_{\text{H}_2\text{SO}_4}=(\text{M}_2\text{V}_2)_{\text{NaOH}}$

$\Rightarrow2\times18.4\text{M}\times100\text{ml}=0.1\text{M}\times\text{V}_2$

$\Rightarrow\text{V}_2=\frac{2\times18.4\times100}{0.1}$

$=36.80\times10^3\text{ml}=36.8\text{L}.$

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